Solve a Nonlinear PDE

I want to solve the problem $$x_1u_{x_1} + 2x_2u_{x_2} + u_{x_3}=3u,~~~~~ u(x_1,x_2,0) = g(x_1,x_2).$$

I believe I am on the right track but would really like feedback. This is what I have so far:

First, let $$\frac{dx_1}{ds} = x_1, ~~~~~ \frac{dx_2}{ds} = 2x_2, ~~~~~ \frac{dx_3}{ds} = 1, ~~~~~\text{and}~~~~~ \frac{dz}{ds} = 3z.$$ Second, let $$x_1(0) = x_1^0,~~~~~ x_2(0) = x_2^0,~~~~~ x_3(0) = 0,~~~~~ \text{and}~~~~~ z(0) = g(x_1^0,x_2^0).$$ Third, from these we can see that we have $$x_1(s) = x_1^0e^s,~~~~~ x_2(s) = x_2^0e^{2s},~~~~~ x_3(s) = s,~~~~~\text{and}~~~~~ z(s) = g(x_1^0,x_2^0)e^{3s}.$$ Fourth, we now see that $$s = x_3,~~~~~ x_1^0 = x_1e^{-x_3},~~~~~\text{and}~~~~~ x_2^0 = x_2e^{-2x_3}.$$ Finally, from this we get $$u(x_1,x_2,x_3) = g\left( x_1e^{-x_3},x_2e^{-2x_3} \right)e^{3x_3}.$$

In order to check if your solution $$(1)$$ is correct, put it into the PDE and see if it agrees or not. $$u(x_1,x_2,x_3) = g\left( x_1e^{-x_3},x_2e^{-2x_3} \right)e^{3x_3}\tag 1$$ Let $$X=x_1e^{-x_3}$$ and $$Y=x_2e^{-2x_3}$$ $$u_{x_1}=\left(e^{-x_3}\frac{\partial g}{\partial X}\right)e^{3x_3}= e^{2x_3}\frac{\partial g}{\partial X}$$ $$u_{x_2}=\left(e^{-2x_3}\frac{\partial g}{\partial Y}\right)e^{3x_3}=e^{x_3}\frac{\partial g}{\partial Y}$$ $$u_{x_3}=3e^{3x_3}g+\left(-x_1e^{-x_3}\frac{\partial g}{\partial X}\right)e^{3x_3}+\left(-2x_2e^{-2x_3}\frac{\partial g}{\partial Y}\right)e^{3x_3}$$ $$u_{x_3}=3e^{3x_3}g-x_1e^{2x_3}\frac{\partial g}{\partial X}-2x_2e^{x_3}\frac{\partial g}{\partial Y}$$ Putting them into the PDE : $$x_1u_{x_1}+2x_2u_{x_2}+u_{x_3}=$$ $$=x_1e^{2x_3}\frac{\partial g}{\partial X}+2x_2e^{x_3}\frac{\partial g}{\partial Y}+3e^{3x_3}g-x_1e^{2x_3}\frac{\partial g}{\partial X}-2x_2e^{x_3}\frac{\partial g}{\partial Y}$$ After simplification : $$x_1u_{x_1}+2x_2u_{x_2}+u_{x_3}=3e^{3x_3}g=3u$$ Thus the PDE is satisfied.
Then checking the condition : $$u(x_1,x_2,0)=g\left( x_1e^{-0},x_2e^{-2(0)} \right)e^{3(0)}=g(x_1,x_2)$$ Thus the condition is satisfied.
Conclusion : Your result $$(1)$$ is a correct solution of the problem.